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\begin{document}

\title{Complex Analysis}
\subtitle{Chapter 2. Complex Functions \\
Section 1. Introduction to the Concept of Analytic Function }
%\institute{SLUC}
\author{LVA}
%\date
%\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
%\date{ {2023年9月21日} }

\maketitle

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\begin{frame}{Contents 1-2}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item  {\color{red}Introduction to the Concept of Analytic Function}
\begin{enumerate}
\item[1.1.] {\color{red}Limits and Continuity}
\item[1.2.] {\color{red}Analytic Functions}
\item[1.3.] {\color{red}Polynomials}
\item[1.4.] {\color{red}Rational Functions}
\end{enumerate}

\item Elementary Theory of Power Series
\begin{enumerate}
\item[2.1.] Sequences
\item[2.2.] Series
\item[2.3.] Uniform Convergence
\item[2.4.] Power Series
\item[2.5.] Abel's Limit Theorem
\end{enumerate}

\end{enumerate}

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\begin{frame}{Contents 3}

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\begin{itemize}

\item[3.] The Exponential and Trigonometric Functions
\begin{enumerate}
\item[3.1.] The Exponential
\item[3.2.] The Trigonometric Functions
\item[3.3.] The Periodicity.
\item[3.4.] The Logarithm
\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1. Introuction to the Concept of Analytic Function }

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\begin{enumerate}
\item[1.] 
The theory of functions of a complex variable aims at extending calculus to the complex domain. 

\item[2.] 
{\color{red}Both differentiation and integration acquire new depth and significance; at the same time the range of applicability becomes radically restricted. }

\item[3.] 
Indeed, only the analytic or holomorphic functions can be freely differentiated and integrated. 

\item[4.] 
They are the only true ``functions'' in the sense of the French ``Theorie des fonctions'' or the German ``Funktionentheorie''.

\end{enumerate}

\end{frame}

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\begin{enumerate}

\item[5.] 
Nevertheless, we shall use the term ``function'' in its modern meaning.

\item[6.] 
Therefore, when stepping up to complex numbers we have to consider {\color{blue}four different kinds of functions}: real functions of a real variable, real functions of a complex variable, complex functions of a real variable, and complex functions of a complex variable.  

\item[7.] 
As a practical matter we agree that the letters $z$ and $w$ shall always denote complex variables; thus, to indicate a complex function of a complex variable we use the notation $w = f(z)$. 

\item[8.] 
The notation $y = f(x)$ will be used in a neutral manner with the understanding that $x$ andy can be either real or complex. 


\end{enumerate}

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\begin{enumerate}

\item[9.] 
When we want to indicate that a variable is definitely restricted to real values, we shall usually denote it by $t$.

\item[10.] 
By these agreements we do not wish to cancel the earlier convention whereby a notation $z = x + iy$ automatically implies that $x$ and $y$ are real. 

\item[11.] 
It is essential that the law by which a function is defined be formulated in clear and unambiguous terms. 

\item[12.] 
In other words, all functions must be well defined and consequently, until further notice, single-valued.  

\item[12a.] 
We shall sometimes use the {\color{blue}pleonastic} term {\it single-valued function} to underline that the function has only one value for each value of the variable.


\end{enumerate}

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\begin{enumerate}

\item[13.] 
It is not necessary that a function be defined for all values of the independent variable.

\item[14.] 
For the moment we shall deliberately under-emphasize the role of point set theory.

\item[15.] 
{\color{red}Therefore we make merely an informal agreement that every function be defined on an open set, by which we mean that if $f(a)$ is defined, then $f(x)$ is defined for all $x$ sufficiently close to $a$. } 

\item[16.] 
The formal treatment of point set topology is {\color{blue}deferred} until the next chapter.

\end{enumerate}

\end{frame}

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\begin{frame}{1.1. Limits and Continuity. }

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\begin{enumerate}

\item[1.]  
The following basic definition will be adopted. 

\item[2.]  
{\color{red} Definition 1. The function $f(x)$ is said to have the limit $A$ as $x$ tends to $a$, 
\begin{equation}
\lim\limits_{x\to a} f(x) = A,
\label{eq-1}
\end{equation}
if and only if the following is true: For every $\varepsilon > 0$ there exists a number $\delta > 0$ with the property that $|f(x) - A| < \delta$ for all values of $x$ such that $|x - a| < \varepsilon$ and $x \neq a$.
}

\item[3.]  
This definition makes decisive use of the absolute value. 

\item[4.]  
Since the notion of absolute value has a meaning for complex as well as for real numbers, we can use the same definition regardless of whether the variable $x$ and the function $f(x)$ are real or complex. 

\end{enumerate}

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\begin{enumerate}

\item[5.]  
As an alternative simpler notation we sometimes write: $f(x)\to A$ for $x\to a$.

\item[6.]  
There are some familiar variants of the definition which correspond to the case where $a$ or $A$ is infinite. 

\item[7.]  
{\color{red}In the real case we can distinguish between the limits $+\infty$ and $-\infty$, but in the complex case there is only one infinite limit. } 

\item[8.]  
We trust the reader to formulate correct definitions to cover all the possibilities. 

\item[9.]  
The well-known results concerning the limit of a sum, a product, and a quotient continue to hold in the complex case. 

\end{enumerate}

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\begin{enumerate}

\item[10.]  
Indeed, the proofs depend only on the properties of the absolute value expressed by 
\begin{equation*}
|ab| = |a|\cdot |b|
\,\,\mathrm{and}\,\,
|a + b| \le |a| + |b|.
\end{equation*}

\item[11.]  
Condition (\ref{eq-1}) is evidently equivalent to
\begin{equation}
\lim\limits_{x\to a} \overline{f(x)} = \overline{A}. 
\label{eq-2}
\end{equation}

\item[12.]  
From (\ref{eq-1}) and (\ref{eq-2}) we obtain
\begin{equation}
\begin{aligned}
\lim\limits_{x\to a} \mathrm{Re}\, f(x) &= \mathrm{Re}\, A\\
\lim\limits_{x\to a} \mathrm{Im}\, f(x) &= \mathrm{Im}\, A. 
\end{aligned}
\label{eq-3}
\end{equation}

\item[13.]  
Conversely, (\ref{eq-1}) is a consequence of (\ref{eq-3}).

\end{enumerate}

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\begin{enumerate}

\item[14.]  
The function $f(x)$ is said to be {\color{blue}continuous} at $a$ if and only if $$\lim\limits_{x\to a} f(x) = f(a).$$

\item[15.]  
A {\color{blue}continuous function}, without further qualification, is one which is continuous at all points where it is defined. 

\item[16.]  
The sum $f(x) + g(x)$ and the product $f(x)g(x)$ of two continuous functions are continuous; the quotient $f(x)/g(x)$ is defined and continuous at $a$ if and only if $g(a) \neq 0$. 

\item[17.]  
If $f(x)$ is continuous, so are $\mathrm{Re}\, f(x)$, $\mathrm{Im}\, f(x)$, and $|f(x)|$. 

\end{enumerate}

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\begin{enumerate}

\item[18.]  
{\color{red}The derivative of a function is defined as a particular limit and can be considered regardless of whether the variables are real or complex. 
}

\item[19.]  
{\color{red}The formal definition is
\begin{equation}
f'(a) = \lim\limits_{x\to a} \frac{f(x)-f(a)}{x-a}.
\label{eq-4}
\end{equation}
}

\item[20.]  
The usual rules for forming the derivative of a sum, a product, or a quotient are all valid. 

\item[21.]  
The derivative of a composite function is determined by the chain rule.

\item[22.]  
There is nevertheless a fundamental difference between the cases of a real and a complex independent variable. 

\end{enumerate}

\end{frame}

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\begin{enumerate}

\item[23.]  
To illustrate our point, let $f(z)$ be a real function of a complex variable whose derivative exists at $z = a$. 

\item[24.]  
Then $f'(a)$ is on one side real, for it is the limit of the quotients 
\begin{equation*}
\frac{f(a + h) - f(a)}{h}
%\label{eq-}
\end{equation*}
as $h$ tends to zero through real values. 

\item[25.]  
On the other side it is also the limit of the quotients
\begin{equation*}
\frac{f(a + ih) - f(a)}{ih}
%\label{eq-}
\end{equation*}
and as such purely imaginary.

\item[26.]  
Therefore $f'(a)$ must be zero. 

\end{enumerate}

\end{frame}

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\begin{enumerate}

\item[27.]  
{\color{red} Thus a real function of a complex variable either has the derivative zero, or else the derivative does not exist. } 

\item[28.]  
The case of a complex function of a real variable can be reduced to the real case. 

\item[29.]  
If we write $z(t) = x(t) + iy(t)$ we find indeed $z'(t) = x'(t) + iy'(t)$, and the existence of $z'(t)$ is equivalent to the simultaneous existence of $x'(t)$ and $y'(t)$.

\end{enumerate}

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\begin{enumerate}

\item[30.]  
The complex notation has nevertheless certain formal advantages which it would be unwise to give up. 

\item[31.]  
{\color{red}In contrast, the existence of the derivative of a complex function of a complex variable has far-reaching consequences for the structural properties of the function. }

\item[32.]  
{\color{red}The investigation of these consequences is the central theme in complex-function theory. } 





\end{enumerate}

\end{frame}

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\begin{frame}{1.2. Analytic Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.]  
The class of {\color{blue}analytic functions} is formed by the complex functions of a complex variable which possess a derivative wherever the function is defined. 

\item[2.]  
The term {\color{blue}holomorphic function} is used with identical meaning. 

\item[3.]  
For the purpose of this preliminary investigation the reader may think primarily of functions which are defined in the whole plane.

\end{enumerate}

\end{frame}

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\begin{enumerate}

\item[4.]  
The sum and the product of two analytic functions are again analytic. 

\item[5.]  
The same is true of the quotient $f(z)/g(z)$ of two analytic functions, provided that $g(z)$ does not vanish. 

\item[6.]  
In the general case it is necessary to exclude the points at which $g(z) = 0$.

\item[7.]  
Strictly speaking, this very typical case will thus not be included in our considerations, but it will be clear that the results remain valid except for obvious modifications.

\end{enumerate}

\end{frame}

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\begin{enumerate}

\item[8.]  
The definition of the derivative can be rewritten in the form
\begin{equation*}
f'(z) = \lim\limits_{h\to 0} \frac{f(z + h) - f(z)}{h}.
\end{equation*}

\item[9.]  
{\color{red}As a first consequence $f(z)$ is necessarily continuous. } 

\item[10.]  
Indeed, from $f(z + h) - f(z) = h \cdot (f(z+h) - f(z))/h$ we obtain
\begin{equation*}
\lim\limits_{h\to 0} (f(z+h)-f(z)) = 0\cdot f'(z) = 0.
\end{equation*}

\item[11.]  
If we write $f(z) = u(z) + iv(z)$ it follows, moreover, that $u(z)$ and $v(z)$ are both continuous.

\end{enumerate}

\end{frame}

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\begin{enumerate}

\item[12.]  
The limit of the difference quotient must be the same regardless of the way in which $h$ approaches zero. 

\item[13.]  
If we choose real values for $h$, then the imaginary part $y$ is kept constant, and the derivative becomes a partial derivative with respect to $x$.

\item[14.]  
We have thus
\begin{equation*}
f'(z) = \frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}.
\end{equation*}

\item[15.]  
Similarly, if we substitute purely imaginary values $ik$ for $h$, we obtain 
\begin{equation*}
f'(z) = \lim\limits_{k\to 0} \frac{f(z+ik)-f(z)}{ik}
= -i \frac{\partial f}{\partial y} 
= -i \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}. 
\end{equation*}

\end{enumerate}

\end{frame}

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\begin{enumerate}

\item[16.]  
It follows that $f(z)$ must satisfy the partial differential equation
\begin{equation}
\frac{\partial f}{\partial x} = -i \frac{\partial f}{\partial y} 
\label{eq-5}
\end{equation}
which resolves into the real equations
\begin{equation}
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \,\, 
\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}. 
\label{eq-6}
\end{equation}

\item[17.]  
These are the {\color{blue}Cauchy-Riemann differential equations} which must be satisfied by the real and imaginary part of any analytic function. 

\end{enumerate}

\end{frame}

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\begin{enumerate}

\item[18.]  
We remark that the existence of the four partial derivatives in (\ref{eq-6}) is
implied by the existence of $f'(z)$. 

\item[19.]  
Using (\ref{eq-6}) we can write down four formally different expressions for $f'(z)$; the simplest is 
\begin{equation*}
f'(z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}
\end{equation*}

\end{enumerate}

\end{frame}

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\begin{enumerate}

\item[20.]  
For the quantity $|f'(z)|^2$ we have, for instance,
\begin{equation*}
|f'(z)|^2 = \left(\frac{\partial u}{\partial x}\right)^2 + 
\left(\frac{\partial u}{\partial y}\right)^2
= \left(\frac{\partial u}{\partial x}\right)^2 + 
\left(\frac{\partial v}{\partial x}\right)^2
=\frac{\partial u}{\partial x}\frac{\partial v}{\partial y} 
- \frac{\partial u}{\partial y}\frac{\partial v}{\partial x}. 
\end{equation*}

\item[21.]  
{\color{red}The last expression shows that $|f'(z)|^2$ is the Jacobian of $u$ and $v$ with respect to $x$ and $y$. } 

\end{enumerate}

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\begin{enumerate}

\item[22.]  
{\color{red}We shall prove later that the derivative of an analytic function is itself analytic. }

\item[23.]  
By this fact $u$ and $v$ will have continuous partial derivatives of all orders, and in particular the mixed derivatives will be equal. 

\item[24.]  
Using this information we obtain from (6)
\begin{equation*}
\begin{aligned}
\Delta u &= \frac{\partial^2 u}{\partial x^2} 
+ \frac{\partial^2 u}{\partial y^2} =0 \\ 
\Delta v &= \frac{\partial^2 v}{\partial x^2} 
+ \frac{\partial^2 v}{\partial y^2} =0 
\end{aligned}
\end{equation*}


\end{enumerate}

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\begin{enumerate}

\item[25.]  
A function $u$ which satisfies {\color{blue}Laplace's equation} $\Delta u = 0$ is said to be {\color{blue}harmonic}.

\item[26.]  
The real and imaginary part of an analytic function are thus harmonic. 

\item[27.]  
{\color{red}If two harmonic functions $u$ and $v$ satisfy the Cauchy-Riemann equations, then $v$ is said to be the conjugate harmonic function of $u$. } 

\item[28.]  
Actually, $v$ is determined only up to an additive constant, so that the use of the definite article, although traditional, is not quite accurate.

\item[29.]  
In the same sense, $u$ is the conjugate harmonic function of $-v$.

\end{enumerate}

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\begin{enumerate}


\item[30.]  
This is not the place to discuss the weakest conditions of regularity which can be imposed on harmonic functions.

\item[31.]  
{\color{red}We wish to prove, however, that the function $u + iv$ determined by a pair of conjugate harmonic functions is always analytic}, and for this purpose we make the explicit assumption that $u$ and $v$ have continuous first-order partial derivatives. 

\end{enumerate}

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\begin{enumerate}

\item[32.]  
It is proved in calculus, under exactly these regularity conditions, that we can write
\begin{equation*}
\begin{aligned}
u(x + h,y + k) - u(x,y) &= \frac{\partial u}{\partial x} h + \frac{\partial u}{\partial y} k + \varepsilon_1 \\ 
v(x + h,y + k) - v(x,y) &= \frac{\partial v}{\partial x} h + \frac{\partial v}{\partial y} k + \varepsilon_2  
\end{aligned}
\end{equation*}
where the remainders $\varepsilon_1$, $\varepsilon_2$ tend to zero more rapidly than $h + ik$ in the sense that $\varepsilon_1/(h + ik)\to 0$ and $\varepsilon_2/(h + ik)\to 0$ for $h + ik\to 0$. 

\end{enumerate}

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\begin{enumerate}

\item[33.]  
With the notation $f(z) = u(x,y) + iv(x,y)$ we obtain by virtue of the relations (6)
\begin{equation*}
f(z + h + ik) - f(z) = \left( \frac{\partial u}{\partial x} \right) (h + ik)
+ \varepsilon_1 + i \varepsilon_2
\end{equation*}
and hence
\begin{equation*}
\lim\limits_{h+ik\to 0} \frac{f(z + h + ik) - f(z)}{h+ik} 
= \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x}.
\end{equation*}

\item[34.]  
We conclude that $f(z)$ is analytic.

\end{enumerate}

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\begin{enumerate}


\item[35.]  
{\color{red} If $u(x,y)$ and $v(x,y)$ have {\color{blue}continuous first-order partial derivatives} which satisfy the Cauchy-Riemann differential equations, then $f(z) = u(z) + iv(z)$ is analytic with continuous derivative $f'(z)$, and conversely. }


\end{enumerate}

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\begin{enumerate}

\item[36.]  
{\color{red}The conjugate of a harmonic function can be found by integration, and in simple cases the computation can be made explicit.
}

\item[37.]  
For instance, $u = x^2-y^2$ is harmonic and 
\begin{equation*}
\frac{\partial u}{\partial x} = 2x, \,\, 
\frac{\partial u}{\partial y} = -2y. 
%\label{eq-}
\end{equation*}

%$\partial u/\partial x = 2x$, $\partial u/\partial y = -2y$. 

\item[38.]  
The conjugate function must therefore satisfy
\begin{equation*}
\frac{\partial v}{\partial x} = 2y, \,\, 
\frac{\partial v}{\partial y} = 2x. 
%\label{eq-}
\end{equation*}


\end{enumerate}

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\begin{enumerate}

\item[39.]  
From the first equation $v = 2xy + \varphi(y)$, where $\varphi(y)$ is a function of $y$ alone. 

\item[40.]  
Substitution in the second equation yields $\varphi'(y) = 0$.

\item[41.]  
Hence $\varphi(y)$ is a constant, and the most general conjugate function of $x^2 - y^2$ is $2xy + c$ where $c$ is a constant. 

\item[42.]  
Observe that $x^2 - y^2 + 2ixy = z^2$. 

\item[43.]  
The analytic function with the real part $x^2 - y^2$ is hence $z^2 + ic$.

\end{enumerate}

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\begin{enumerate}

\item[44.]  
{\color{red}There is an interesting formal procedure which throws considerable light on the nature of analytic functions.
}

\item[45.]  
We present this procedure with an explicit warning to the reader that it is purely formal and does not possess any power of proof. 

\end{enumerate}

\end{frame}

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\begin{enumerate}

\item[46.]  
Consider a complex function $f(x,y)$ of two real variables. 

\item[47.]  
Introducing the complex variable $z = x + iy$ and its conjugate $\bar{z} = x - iy$, we can write 
$$x = \frac{z + \bar{z}}{2},\,\, y = \frac{z - \bar{z}}{2i}. $$ 

\item[48.]  
With this change of variable we can consider $f(x,y)$ as a function of $z$ and $\bar{z}$ which we will treat as independent variables (forgetting that they are in fact conjugate to each other). 

\item[49.]  
{\color{blue}If the rules of calculus were applicable}, we would obtain
\begin{equation*}
\frac{\partial f}{\partial z} = \frac{1}{2} \left( \frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y} \right), \,\,\,\,\,
\frac{\partial f}{\partial \bar{z}} = \frac{1}{2} \left( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \right). 
%\label{eq-}
\end{equation*}

\end{enumerate}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[50.]  
These expressions have no convenient definition as limits, but we can nevertheless introduce them as symbolic derivatives with respect to $z$ and $\bar{z}$. 

\item[51.]  
{\color{red}By comparison with (\ref{eq-5}) we find that analytic functions are characterized by the condition $$\frac{\partial f}{\partial \bar{z}} = 0.$$
}

\item[52.]  
We are thus tempted to say that an analytic function is independent of $\bar{z}$, and a function of $z$ alone. 

\end{enumerate}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[53.]  
This formal reasoning supports the point of view that analytic functions are true functions of a complex variable as opposed to functions which are more adequately described as complex functions of two real variables.

\end{enumerate}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[54.]  

{\color{red}By similar formal arguments we can derive a very simple method which allows us to compute, without use of integration, the analytic function $f(z)$ whose real part is a given harmonic function $u(x,y)$. 
}


\end{enumerate}

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\begin{frame}{1.2. Analytic Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[55.]  
We remark first that the conjugate function $\overline{f(z)}$ has the derivative zero with respect to $z$ and may, therefore, be considered as a function of $\bar{z}$; we denote this function by $\bar{f}(\bar{z})$.

\item[56.]  
With this notation we can write down the identity
\begin{equation*}
u(x,y) = \frac{1}{2}[f(x + iy) + \bar{f}(x - iy)].
%\label{eq-}
\end{equation*}


\item[57.]  
It is reasonable to expect that this is a formal identity, and then it holds
even when $x$ and $y$ are complex. 

\item[58.]  
If we substitute $x = z/2$, $y = z/2i$, we obtain
\begin{equation*}
u(z/2, z/2i) = \frac{1}{2}[f(z) + \bar{f}(0)].
%\label{eq-}
\end{equation*}

\end{enumerate}

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\begin{frame}{1.2. Analytic Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[59.]  
Since $f(z)$ is only determined up to a purely imaginary constant, we may as well assume that $f(0)$ is real, which implies $\bar{f}(0) = u(O,O)$. 

\item[60.]  
The function $f(z)$ can thus be computed by means of the formula 
\begin{equation*}
f(z) = 2u(z/2, z/2i)-u(0,0).
%\label{eq-}
\end{equation*}

\item[61.]  
A purely imaginary constant can be added at will.

\end{enumerate}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[62.]  
{\color{red}In this form the method is definitely limited to functions $u(x,y)$ which are rational in $x$ and $y$, for the function must have a meaning for complex values of the argument.}

\item[63.]  
Suffice it to say that the method can be extended to the general case and that a complete justification can be given.

\end{enumerate}

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\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $g(w)$ and $f(z)$ are analytic functions, show that $g(f(z))$ is also analytic.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

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\begin{frame}{1.2. Analytic Functions. Exercise - 2 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Verify Cauchy-Riemann's equations for the functions $z^2$ and $z^3$. 
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.2. Analytic Functions. Exercise - 3 \hfill 作业2A-5 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Find the most general harmonic polynomial of the form $ax^3+bx^2y+cxy^2+dy^3$. 
Determine the conjugate harmonic function and the corresponding analytic function by integration and by the formal method.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

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\begin{frame}{1.2. Analytic Functions. Exercise - 4}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that an analytic function cannot have a constant absolute value without reducing to a constant. 
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.2. Analytic Functions. Exercise - 5}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove rigorously that the functions $f(z)$ and $\overline{f(\bar{z})}$ are simultaneously analytic.
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.2. Analytic Functions. Exercise - 6}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Prove that the functions $u(z)$ and $u(\bar{z})$ are simultaneously harmonic. 
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.2. Analytic Functions. Exercise - 7}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Show that a harmonic function satisfies the formal differential equation
$$
\frac{\partial^2 u}{\partial z \partial \bar{z}} =0. 
$$
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.3. Polynomials.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[1.] 
Every constant is an analytic function with the derivative 0.

\item[2.] 
The simplest nonconstant analytic function is $z$ whose derivative is 1.

\item[3.] 
Since {\color{blue}the sum and product of two analytic functions are again analytic}, it follows that every polynomial
\begin{equation}
P(z) = a_0 + a_1z + \cdots + a_nz^n
\label{eq-7}
\end{equation}
is an analytic function.

\item[4.] 
Its derivative is
\begin{equation*}
P'(z) = a_1 + 2a_2z + \cdots + na_nz^{n-1}.
\end{equation*}

\end{enumerate}

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\begin{frame}{1.3. Polynomials.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[5.] 
The notation (\ref{eq-7}) shall imply that $a_n \neq 0$, and {\color{blue}the polynomial is then said to be of degree $n$}. 

\item[6.] 
The constant 0, considered as a polynomial, is in many respects exceptional and will be excluded from our considerations. 

\item[7.] 
For formal reasons, if the constant 0 is regarded as a polynomial, its degree is set equal to $-\infty$. 

\item[8.] 
For $n > 0$ the equation $P(z) = 0$ has at least one root. 

\item[9.] 
This is the so-called {\color{red}fundamental theorem of algebra} which we shall prove later. 

\end{enumerate}

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\begin{frame}{1.3. Polynomials.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[10.] 
If $P(\alpha_1) = 0$, it is shown in elementary algebra that $P(z) = (z-\alpha_1)P_1(z)$ where $P_1(z)$ is a polynomial of degree $n-1$. 

\item[11.] 
Repetition of this process finally leads to a complete factorization 
\begin{equation}
P(z) = a_n(z-\alpha_1)(z-\alpha_2)\cdots (z-\alpha_n)
\label{eq-8}
\end{equation}
where the $\alpha_1, \alpha_2,\cdots, \alpha_n$ are not necessarily distinct. 

\item[12.] 
From the factorization we conclude that $P(z)$ does not vanish for any value of $z$ 
different from $\alpha_1, \alpha_2, \cdots, \alpha_n$.

\item[13.] 
Moreover, the factorization is uniquely determined except for the order of the factors.

\end{enumerate}

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\begin{frame}{1.3. Polynomials.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[14.] 
If exactly $h$ of the $\alpha_i$ coincide, their common value is called {\color{blue}a zero of $P(z)$ of the order $h$}.

\item[15.] 
We find that the sum of the orders of the zeros of a polynomial is equal to its degree.

\item[16.] 
More simply, if each zero is counted as many times as its order indicates, a polynomial of degree $n$ has exactly $n$ zeros.

\end{enumerate}

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\begin{frame}{1.3. Polynomials.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[17.] 
{\color{red}The order of a zero $\alpha$ can also be determined by consideration of the successive derivatives of $P(z)$ for $z = a$. } 

\item[18.] 
Suppose that $\alpha$ is a zero of order $h$.

\item[19.] 
Then we can write $P(z) = (z-\alpha)^hP_h(z)$ with $P_h(\alpha)\neq 0$. 

\item[20.] 
Successive derivation yields $P(\alpha) = P'(\alpha) = \cdots = P^{(h-1)}(\alpha) = 0$ while $P^{(h)}(\alpha) \neq 0$.

\item[21.] 
In other words, {\color{red}the order of a zero equals the order of the first nonvanishing derivative. }

\item[22.] 
A zero of order 1 is called a {\color{blue}simple zero} and is characterized by the conditions $P(\alpha) = 0$, $P'(\alpha) \neq 0$.

\end{enumerate}

\end{frame}

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\begin{frame}{1.3. Polynomials. \hfill 作业2A-6 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[23.] 
As an application we shall prove the following theorem, known as {\color{blue}Lucas's theorem}. 

\item[24.] 
{\color{red}Theorem 1. If all zeros of a polynomial $P(z)$ lie in a half plane, then all zeros of the derivative $P'(z)$ lie in the same half plane.
}

\item[25.] 
From %(\ref{eq-8}) 
\begin{equation*}
P(z) = a_n(z-\alpha_1)(z-\alpha_2)\cdots (z-\alpha_n)
%\label{eq-8}
\end{equation*}
we obtain
\begin{equation}
\frac{P'(z)}{P(z)} = \frac{1}{z-\alpha_1} + \cdots + \frac{1}{z-\alpha_n}. 
\label{eq-9}
\end{equation}

\end{enumerate}

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\begin{frame}{1.3. Polynomials.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[26.] 
{\color{blue}Suppose that the half plane $H$ is defined as} the part of the plane where 
$$\mathrm{Im}\, \frac{z-a}{b} < 0. $$ (See Chap. 1, Sec. 2.3). 

\item[27.] 
{\color{blue}If $\alpha_k$ is in $H$ and $z$ is not}, we have then
\begin{equation*}
\mathrm{Im}\, \frac{z-\alpha_k}{b} = \mathrm{Im}\, \frac{z-a}{b} - \mathrm{Im}\, \frac{\alpha_k -a}{b} > 0. 
\end{equation*}

\item[28.] 
But the imaginary parts of reciprocal numbers have opposite sign. 

\item[29.] 
Therefore, under the same assumption, 
$$\mathrm{Im}\, \frac{b}{z - \alpha_k} < 0. $$

\end{enumerate}

\end{frame}

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\begin{frame}{1.3. Polynomials.  }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[30.] 
If this is true for all $k$ we conclude from %(\ref{eq-9}) 
\begin{equation*}
\frac{P'(z)}{P(z)} = \frac{1}{z-\alpha_1} + \cdots + \frac{1}{z-\alpha_n} 
%\label{eq-9}
\end{equation*}
that
\begin{equation*}
\mathrm{Im}\, \frac{bP'(z)}{P(z)}
= \sum\limits_{k=1}^{n} \mathrm{Im}\, \frac{b}{z-\alpha_k} < 0, 
\end{equation*}
and consequently $P'(z) \neq 0$.

\item[31.] 
{\color{red}In a sharper formulation the theorem tells us that the smallest convex polygon that contains the zeros of $P(z)$ also contains the zeros of $P'(z)$. } 


\end{enumerate}

\end{frame}

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\begin{frame}{1.4. Rational Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}
\item[1.] 
We turn to the case of a rational function
\begin{equation}
R(z) = \frac{P(z)}{Q(z)},
\label{eq-10}
\end{equation}
given as the quotient of two polynomials. 

\item[2.] 
We assume, and this is essential, that $P(z)$ and $Q(z)$ have no common factors and hence no common zeros. 

\item[3.] 
$R(z)$ will be given the value $\infty$ at the zeros of $Q(z)$. 

\item[4.] 
It must therefore be considered as a function {\color{blue}with values in the extended plane}, and as such it is continuous. 

\item[5.] 
The zeros of $Q(z)$ are called {\color{blue}poles} of $R(z)$, and the {\color{blue}order of a pole} is by definition equal to the order of the corresponding zero of $Q(z)$.

 
\end{enumerate}

\end{frame}

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\begin{frame}{1.4. Rational Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[6.] 
The derivative
\begin{equation}
R'(z) = \frac{P'(z)Q(z)-Q'(z)P(z)}{Q(z)^2}
\label{eq-11}
\end{equation}
exists only when $Q(z) \neq 0$.

\item[7.] 
However, as a rational function defined by the right-hand member of (\ref{eq-11}), $R'(z)$ has the same poles as $R(z)$, the order of each pole being increased by one. 

\item[8.] 
In case $Q(z)$ has multiple zeros, it should be noticed that the expression (\ref{eq-11}) does not appear in reduced form.

 
\end{enumerate}

\end{frame}

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\begin{frame}{1.4. Rational Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[9.] 
{\color{red}Greater unity is achieved if we let the variable $z$ as well as the values $R(z)$ range over the extended plane. } 

\item[10.] 
We may define $R(\infty)$ as the limit of $R(z)$ as $z\to\infty$, but this definition would not determine the order of a zero or pole at $\infty$.

\item[11.] 
It is therefore preferable to consider the function $R(l/z)$, which we can rewrite as a rational function $R_1(z)$, and set
\begin{equation*}
R(\infty) = R_1(0).
%\label{eq-}
\end{equation*}

\item[12.] 
If $R_1(0) = 0$ or $\infty$, the order of the zero or pole at $\infty$ is defined as the order of the zero or pole of $R_1(z)$ at the origin. 
 
\end{enumerate}

\end{frame}

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\begin{frame}{1.4. Rational Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[13.] 
With the notation
\begin{equation*}
R(z) = \frac{a_0 + a_1z + \cdots + a_nz^n}{b_0 + b_1z + \cdots + b_mz^m}
%\label{eq-}
\end{equation*}
we obtain
\begin{equation*}
R_1(z) = z^{m-n}\frac{a_0z^n + a_1z^{n-1} + \cdots + a_n}{b_0z^m + b_1z^{m-1} + \cdots + b_m}
%\label{eq-}
\end{equation*}
where the power $z^{m-n}$ belongs either to the numerator or to the denominator. 

\item[14.] 
Accordingly, if $m > n$ the rational function $R(z)$ has a zero of order $m - n$ at $\infty$, if $m < n$ the point at $\infty$ is a pole of order $n-m$, and if $m = n$ 
\begin{equation*}
R(\infty) = \frac{a_n}{b_m} \neq 0, \infty. 
%\label{eq-}
\end{equation*}

 
\end{enumerate}

\end{frame}

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\begin{frame}{1.4. Rational Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[15.] 
{\color{red}We can now count the total number of zeros and poles in the extended plane. } 

\item[16.] 
The count shows that the number of zeros, including those at $\infty$, is equal to the greater of the numbers $m$ and $n$. 

\item[17.] 
The number of poles is the same.

\item[18.] 
This common number of zeros and poles is called the {\color{blue}order of the rational function}.

 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vfill 

%\vspace{-0.4cm}
\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

{\footnotesize 
For example, the order of the rational function $$f(z) = \frac{z^2+1}{z^3+1}$$ 
is three. It has three zeros $\{i,-i,\infty\}$ and three poles $\{-1,\omega, \omega^2\}$. 
}

\end{enumerate}

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\begin{frame}{1.4. Rational Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}


\item[19.] 
If $a$ is any constant, the function $R(z) - a$ has the same poles as $R(z)$, and consequently the same order. 

\item[20.] 
The zeros of $R(z) - a$ are roots of the equation $R(z) = a$, and if the roots are counted as many times as the order of the zero indicates, we can state the following result.

\item[21.] 
{\color{red}A rational function $R(z)$ of order $p$ has $p$ zeros and $p$ poles, and every equation $R(z) = a$ has exactly $p$ roots.
}

\end{enumerate}

\end{frame}

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\begin{frame}{1.4. Rational Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[22.] 
A rational function of order 1 is a linear fraction
\begin{equation*}
S(z) = \frac{\alpha z + \beta}{\gamma z + \delta}
%\label{eq-}
\end{equation*}
with $\alpha\delta - \beta\gamma \neq 0$. 

\item[23.] 
Such fractions, or linear transformations, will be studied at length in Chap. 3, Sec. 3. 

\item[24.] 
For the moment we note merely that the equation $w = S(z)$ has exactly one root, and we find indeed 
\begin{equation*}
z = S^{-1}(w) = \frac{\delta w - \beta}{-\gamma w + \alpha}. 
%\label{eq-}
\end{equation*}

\end{enumerate}

\end{frame}

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\begin{frame}{1.4. Rational Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[25.] 
The transformations $S$ and $S^{-1}$ are inverse to each other. 

\item[26.] 
The linear transformation $z+a$ is called a {\color{blue}parallel translation}, and $1/z$ is an {\color{blue}inversion}. 

\item[27.] 
The former has a fixed point at $\infty$, the latter interchanges $0$ and $\infty$.

\end{enumerate}

\end{frame}

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\begin{frame}{1.4. Rational Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[28.] 
{\color{red}Every rational function has a representation by partial fractions. }

\item[29.] 
In order to derive this representation we assume first that $R(z)$ has a pole at $\infty$. 

\item[30.] 
We carry out the division of $P(z)$ by $Q(z)$ until the degree of the remainder is at most equal to that of the denominator. 

\item[31.] 
The result can be written in the form
\begin{equation}
R(z) = G(z) + H(z)
\label{eq-12}
\end{equation}
where $G(z)$ is a polynomial {\color{orange}without constant term}, and {\color{blue}$H(z)$ is finite at $\infty$}.

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\vfill 

%\vspace{-0.4cm}
\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

{\footnotesize 
For example, 
$$R(z) = \frac{z^5+z^4+z^3+z^2+z+2}{z^2+2z+1} 
= z^3 - z^2 + 2z + \frac{-2z^2-z+2}{z^2+2z+1}
$$

}


\end{enumerate}

\end{frame}

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\begin{frame}{1.4. Rational Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[32.] 
The degree of $G(z)$ is the order of the pole at $\infty$, and the polynomial $G(z)$ is called {\color{red}the singular part of $R(z)$ at $\infty$}.

\item[33.] 
Let the distinct finite poles of $R(z)$ be denoted by $\beta_1, \beta_2, \cdots, \beta_q$. 

\item[34.] 
The function $R \left(\beta_j + \frac{1}{\zeta} \right)$ is a rational function of $\zeta$ with a pole at $\zeta = \infty$.

\item[35.] 
By use of the decomposition (\ref{eq-12}) we can write
\begin{equation*}
R\left(\beta_j + \frac{1}{\zeta} \right) = G_j(\zeta) + H_j(\zeta),
%\label{eq-}
\end{equation*}
or with a change of variable
\begin{equation*}
R(z) = G_j\left(\frac{1}{z-\beta_j} \right)
+ H_j\left(\frac{1}{z-\beta_j} \right) 
%\label{eq-}
\end{equation*}
 

\end{enumerate}

\end{frame}

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\begin{frame}{1.4. Rational Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[36.] 
Here $G_j\left(\frac{1}{z-\beta_j} \right)$ is a polynomial in $\frac{1}{z-\beta_j}$; without constant term, called {\color{red}the singular part of $R(z)$ at $\beta_j$}. 

\item[37.] 
The function $H_j\left(\frac{1}{z-\beta_j} \right)$ is finite for $z=\beta_j$.

\item[38.] 
Consider now the expression
\begin{equation}
R(z) - G(z) - \sum\limits_{j=1}^{q} G_j\left(\frac{1}{z-\beta_j} \right).
\label{eq-13}
\end{equation}

\item[39.] 
This is a rational function which cannot have other poles than $\beta_1, \beta_2,
\cdots, \beta_q$ and $\infty$.
 

\end{enumerate}

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\begin{frame}{1.4. Rational Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[40.] 
At $z = \beta_j$ we find that the two terms which become infinite have a difference $H_j\left( \frac{1}{z-\beta_j}\right)$ with a finite limit, and the same is true at $\infty$.

\item[41.] 
Therefore (\ref{eq-13}) has neither any finite poles nor a pole at $\infty$.

\item[42.] 
{\color{red}A rational function without poles must reduce to a constant, and if this constant is absorbed in $G(z)$ we obtain 
\begin{equation}
R(z) = G(z) + \sum\limits_{i=1}^{q} G_j \left( \frac{1}{z-\beta_j} \right).
\label{eq-14}
\end{equation}
}

\end{enumerate}

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\begin{frame}{1.4. Rational Functions. }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{enumerate}

\item[43.] 
This representation is well known from the calculus where it is used as a technical device in integration theory.

\item[44.] 
However, it is only with the introduction of complex numbers that it becomes completely successful. 

\end{enumerate}

\end{frame}

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\begin{frame}{1.4. Rational Functions. Exercise - 1 \hfill 作业2A-7}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Use the method of the text to develop
$$
\frac{z^4}{z^3-1} \,\,\,\,\mathrm{and}\,\,\,\, \frac{1}{z(z+1)^2(z+2)^3}
$$
in partial fractions.
}

\item  Answer. 
\begin{enumerate}
\item 
$$
\frac{z^4}{z^3-1} = z + \frac{1}{3}\frac{1}{z-1}
 + \frac{1}{3\omega}\frac{1}{z-\omega}
 + \frac{1}{3\omega^2}\frac{1}{z-\omega^2}.
$$

 
\item 


\end{enumerate}

\end{itemize}

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\begin{frame}{1.4. Rational Functions. Exercise - 2 \hfill 作业2A-8 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $Q$ is a polynomial with distinct roots $\alpha_1,\cdots,\alpha_n$, and if $P$ is a polynomial of degree $<n$, show that 
$$
\frac{P(z)}{Q(z)} = \sum\limits_{k=1}^{n} \frac{P(\alpha_k)}{Q'(\alpha_k)(z-\alpha_k)}. 
$$
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.4. Rational Functions. Exercise - 3}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
Use the formula in the preceding exercise to prove that there exists a unique polynomial $P$ of degree $<n$ with given values $c_k$ at the points $\alpha_k$ (Lagrange's interpolation polynomial).
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.4. Rational Functions. Exercise - 4}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
What is the general form of a rational function which has absolute value 1 on the circle $|z| = 1$? In particular, how are the zeros and poles related to each other?
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.4. Rational Functions. Exercise - 5}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If a rational function is real on $|z|=1$, how are the zeros and poles situated?
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}

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\begin{frame}{1.4. Rational Functions. Exercise - 6}

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}

\begin{itemize}
\item  {\color{red}Question. 
If $R(z)$ is a rational function of order $n$, how large and how small can the order of $R'(z)$ be?
}

\item  Answer. 
\begin{enumerate}
\item 
 
\item 

\item 

\end{enumerate}

\end{itemize}

\end{frame}


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